Three isotopes of an element have mass numbers, m, (m + 1) and (m + 2). If the mean mass number is (m + 0.5) then which of the following ratios can be accepted for m, (m +1), (m + 2) in that order ?

- a)1 : 1 : 1
- b)4 : 1 : 1
- c)3 : 2 : 1
- d)2 : 1 : 1

Correct answer is option 'B'. Can you explain this answer?

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Rakshit Dayma
answered
Aug 04, 2018

The mass number is the total number of protons and neutrons in the atom. But this is more of a maths question rather than chemistry.

Let's assume there are x units of the isotope with mass number m

y units of the isotope with mass number m+1

z units of the isotope with mass number m+2

So the mean would be (xm + y(m+1) + z(m+2)) / (x+y+z) [Mean = Total sum/Total quantity]

which is equal to m+0.5 [Given in question]

Solving this we get, 0.5(x - y - 3z) = 0=> x = y+3z -----------equation (1)

Now we can put values of y and z and get x. There can be infinitely many values of x, y and z.

Here amongst the options, we can see that option (b) matches this criterion.

To be precise, if we take a constant k, then x=4k, y=k and c=k satisfy equation (1).

We can also see that option (a) and option (c) don't satisfy equation (1)

So amongst the options, option (b) is the answer.

As a check, if we take other ratios which satisfy equation (1) they can also be an answer to the question.

e.g. 7:1:2 => x=7k, y=k, z=2k

Now the mean would be, (xm + y(m+1) + z(m+2))/(x+y+z)

Putting the values,

(7km + k(m+1) + 2k(m+2))/(7k + k + 2k)

= (10 km + 5k)/(10k)

= m + 0.5

So 7:1:2 can also be a ratio which satisfies the given criterion.

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